id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-23600 | Simple Spectrum Approximation of Diagonalizable Matrix/Property 5 | :There exists $\tau \in \R_{>0}$ such that $0 < t < \tau \implies \map \sigma {\map {\mathbf A} t}$ is simple | Let $\lambda_i = \lambda_j$ be a non-simple eigenvalue of $\mathbf A$.
Let $\varepsilon_i$ be the $i^{th}$ diagonal element of $\mathbf E$.
Then, by a previous proof, the corresponding eigenvalues of $\map {\mathbf A} t$ are
:$\lambda_i + t \varepsilon_i$
:$\lambda_j + t \varepsilon_j$
The difference between these when... | :There exists $\tau \in \R_{>0}$ such that $0 < t < \tau \implies \map \sigma {\map {\mathbf A} t}$ is [[Definition:Simple Eigenvalue|simple]] | Let $\lambda_i = \lambda_j$ be a non-[[Definition:Simple Eigenvalue|simple eigenvalue]] of $\mathbf A$.
Let $\varepsilon_i$ be the $i^{th}$ [[Definition:Diagonal Element|diagonal element]] of $\mathbf E$.
Then, by [[Simple Spectrum Approximation of Diagonalizable Matrix/Property 4|a previous proof]], the correspondi... | Simple Spectrum Approximation of Diagonalizable Matrix/Property 5 | https://proofwiki.org/wiki/Simple_Spectrum_Approximation_of_Diagonalizable_Matrix/Property_5 | https://proofwiki.org/wiki/Simple_Spectrum_Approximation_of_Diagonalizable_Matrix/Property_5 | [
"Simple Spectrum Approximation of Diagonalizable Matrix"
] | [
"Definition:Simple Eigenvalue"
] | [
"Definition:Simple Eigenvalue",
"Definition:Main Diagonal/Diagonal Elements",
"Simple Spectrum Approximation of Diagonalizable Matrix/Property 4",
"Definition:Eigenvalue/Square Matrix",
"Definition:Simple Eigenvalue",
"Definition:Simple Eigenvalue",
"Definition:Absolute Difference",
"Definition:Eigenv... |
proofwiki-23601 | Diagonalizable Linear Operator Restricted to Invariant Subspace is Diagonalizable/Lemma | Let $\set {v_i}$ be a subset of eigenvectors of $A$ each associated with a distinct eigenvalue.
Then:
:$\ds \sum_i v_i \in W \implies \forall i : v_i \in W$ | Proof by induction.
Let $\map P n$ be the statement that:
:If $\size {\set {v_i}} = n$, then $\ds \sum_{i \mathop = 1}^n v_i \in W \implies \forall i \in \closedint 1 n : v_i \in W$ | Let $\set {v_i}$ be a [[Definition:Subset|subset]] of [[Definition:Eigenvector of Linear Operator|eigenvectors]] of $A$ each associated with a distinct [[Definition:Eigenvalue of Linear Operator|eigenvalue]].
Then:
:$\ds \sum_i v_i \in W \implies \forall i : v_i \in W$ | Proof by [[Definition:Principle of Mathematical Induction|induction]].
Let $\map P n$ be the [[Definition:Statement|statement]] that:
:If $\size {\set {v_i}} = n$, then $\ds \sum_{i \mathop = 1}^n v_i \in W \implies \forall i \in \closedint 1 n : v_i \in W$ | Diagonalizable Linear Operator Restricted to Invariant Subspace is Diagonalizable/Lemma | https://proofwiki.org/wiki/Diagonalizable_Linear_Operator_Restricted_to_Invariant_Subspace_is_Diagonalizable/Lemma | https://proofwiki.org/wiki/Diagonalizable_Linear_Operator_Restricted_to_Invariant_Subspace_is_Diagonalizable/Lemma | [
"Diagonalizable Linear Operator Restricted to Invariant Subspace is Diagonalizable"
] | [
"Definition:Subset",
"Definition:Eigenvector/Linear Operator",
"Definition:Eigenvalue/Linear Operator"
] | [
"Principle of Mathematical Induction",
"Definition:Statement"
] |
proofwiki-23602 | Diagonalizable Linear Operator Restricted to Invariant Subspace is Diagonalizable | Let $V$ be a vector space.
Let $A : V \to V$ be a diagonalizable linear operator.
Let $W \subseteq V$ be an invariant subspace of $\mathbf A$.
Let $A {\restriction_W}$ be the restriction of $A$ to $W$.
Then, $A {\restriction_W}$ is diagonalizable in $W$. | === Lemma ===
{{:Diagonalizable Linear Operator Restricted to Invariant Subspace is Diagonalizable/Lemma}}{{qed|lemma}}
Since $A$ is diagonalizable it has an eigenbasis.
Therefore, every vector in $W$ can be written as a linear combination of a subset of the eigenvectors of $A$.
By the {{Lemma|Diagonalizable Linear Ope... | Let $V$ be a [[Definition:Vector Space|vector space]].
Let $A : V \to V$ be a [[Definition:Diagonalizable Linear Operator|diagonalizable linear operator]].
Let $W \subseteq V$ be an [[Definition:Invariant Subspace|invariant subspace]] of $\mathbf A$.
Let $A {\restriction_W}$ be the [[Definition:Restriction of Mappin... | === [[Diagonalizable Linear Operator Restricted to Invariant Subspace is Diagonalizable/Lemma|Lemma]] ===
{{:Diagonalizable Linear Operator Restricted to Invariant Subspace is Diagonalizable/Lemma}}{{qed|lemma}}
Since $A$ is [[Definition:Diagonalizable Linear Operator|diagonalizable]] it has an [[Definition:Eigenbasi... | Diagonalizable Linear Operator Restricted to Invariant Subspace is Diagonalizable | https://proofwiki.org/wiki/Diagonalizable_Linear_Operator_Restricted_to_Invariant_Subspace_is_Diagonalizable | https://proofwiki.org/wiki/Diagonalizable_Linear_Operator_Restricted_to_Invariant_Subspace_is_Diagonalizable | [
"Diagonalizable Linear Operator Restricted to Invariant Subspace is Diagonalizable",
"Diagonalizable Linear Operators",
"Restrictions"
] | [
"Definition:Vector Space",
"Definition:Diagonalizable Linear Operator",
"Definition:Invariant Subspace",
"Definition:Restriction/Mapping",
"Definition:Diagonalizable Linear Operator"
] | [
"Diagonalizable Linear Operator Restricted to Invariant Subspace is Diagonalizable/Lemma",
"Definition:Diagonalizable Linear Operator",
"Definition:Eigenbasis/Vector Space",
"Definition:Vector",
"Definition:Linear Combination",
"Definition:Subset",
"Definition:Eigenvector/Linear Operator",
"Definition... |
proofwiki-23603 | Diagonalizable Matrices Commute iff Simultaneously Diagonalizable | Let $\mathbb F$ be a field.
Let $M$ be a set of diagonalizable matrices over $\mathbb F$.
Then, $M$ is a commuting set {{Iff}} $M$ is simultaneously diagonalizable. | === Necessary Case ===
We have that $M$ is simultaneously diagonalizable.
Thus, there is a nonsingular matrix $\mathbf X$ that diagonalizes all elements of $M$.
Let $\mathbf A, \mathbf B \in M$ be arbitrary.
Let $\mathbf D_{\mathbf A} = \mathbf X^{-1} \mathbf A \mathbf X$
Let $\mathbf D_{\mathbf B} = \mathbf X^{-1} \ma... | Let $\mathbb F$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $M$ be a [[Definition:Set|set]] of [[Definition:Diagonalizable Matrix|diagonalizable matrices]] over $\mathbb F$.
Then, $M$ is a [[Definition:Commuting Set of Elements|commuting set]] {{Iff}} $M$ is [[Definition:Simultaneously Diagonalizable Mat... | === Necessary Case ===
We have that $M$ is [[Definition:Simultaneously Diagonalizable Matrices|simultaneously diagonalizable]].
Thus, there is a [[Definition:Nonsingular Matrix|nonsingular matrix]] $\mathbf X$ that [[Definition:Diagonalizable Matrix|diagonalizes]] all [[Definition:Element|elements]] of $M$.
Let $\m... | Diagonalizable Matrices Commute iff Simultaneously Diagonalizable | https://proofwiki.org/wiki/Diagonalizable_Matrices_Commute_iff_Simultaneously_Diagonalizable | https://proofwiki.org/wiki/Diagonalizable_Matrices_Commute_iff_Simultaneously_Diagonalizable | [
"Diagonalizable Matrices",
"Commutativity"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Set",
"Definition:Diagonalizable Matrix",
"Definition:Commutative/Set",
"Definition:Simultaneously Diagonalizable Matrices"
] | [
"Definition:Simultaneously Diagonalizable Matrices",
"Definition:Nonsingular Matrix",
"Definition:Diagonalizable Matrix",
"Definition:Element",
"Definition:Arbitrary",
"Definition:Diagonal Matrix",
"Matrix Multiplication is Associative",
"Matrix Multiplication on Diagonal Matrices is Commutative",
"... |
proofwiki-23604 | Rule of Association/Conjunction/Formulation 2/Proof 1/Reverse Implication | :$\vdash \paren {p \land \paren {q \land r} } \impliedby \paren {\paren {p \land q} \land r}$ | {{BeginTableau|\vdash \paren {\paren {p \land q} \land r} \implies \paren {p \land \paren {q \land r} } }}
{{Assumption |1|\paren {p \land q} \land r}}
{{Simplification|2|1|p \land q|1|1}}
{{Simplification|3|1|r|1|2}}
{{Simplification|4|1|p|2|1}}
{{Simplification|5|1|q|2|2}}
{{Conjunction|6|1|q \land r|5|3}}
{{Conjunc... | :$\vdash \paren {p \land \paren {q \land r} } \impliedby \paren {\paren {p \land q} \land r}$ | {{BeginTableau|\vdash \paren {\paren {p \land q} \land r} \implies \paren {p \land \paren {q \land r} } }}
{{Assumption |1|\paren {p \land q} \land r}}
{{Simplification|2|1|p \land q|1|1}}
{{Simplification|3|1|r|1|2}}
{{Simplification|4|1|p|2|1}}
{{Simplification|5|1|q|2|2}}
{{Conjunction|6|1|q \land r|5|3}}
{{Conjunc... | Rule of Association/Conjunction/Formulation 2/Proof 1/Reverse Implication | https://proofwiki.org/wiki/Rule_of_Association/Conjunction/Formulation_2/Proof_1/Reverse_Implication | https://proofwiki.org/wiki/Rule_of_Association/Conjunction/Formulation_2/Proof_1/Reverse_Implication | [
"Rule of Association"
] | [] | [] |
proofwiki-23605 | Rule of Association/Conjunction/Formulation 2/Proof 1/Forward Implication | :$\vdash \paren {p \land \paren {q \land r} } \implies \paren {\paren {p \land q} \land r}$ | {{BeginTableau|\vdash \paren {p \land \paren {q \land r} } \implies \paren {\paren {p \land q} \land r} }}
{{Assumption |1|p \land \paren {q \land r} }}
{{Simplification|2|1|p|1|1}}
{{Simplification|3|1|q \land r|1|2}}
{{Simplification|4|1|q|3|1}}
{{Simplification|5|1|r|3|2}}
{{Conjunction|6|1|p \land q|2|4}}
{{Conjun... | :$\vdash \paren {p \land \paren {q \land r} } \implies \paren {\paren {p \land q} \land r}$ | {{BeginTableau|\vdash \paren {p \land \paren {q \land r} } \implies \paren {\paren {p \land q} \land r} }}
{{Assumption |1|p \land \paren {q \land r} }}
{{Simplification|2|1|p|1|1}}
{{Simplification|3|1|q \land r|1|2}}
{{Simplification|4|1|q|3|1}}
{{Simplification|5|1|r|3|2}}
{{Conjunction|6|1|p \land q|2|4}}
{{Conjun... | Rule of Association/Conjunction/Formulation 2/Proof 1/Forward Implication | https://proofwiki.org/wiki/Rule_of_Association/Conjunction/Formulation_2/Proof_1/Forward_Implication | https://proofwiki.org/wiki/Rule_of_Association/Conjunction/Formulation_2/Proof_1/Forward_Implication | [
"Rule of Association"
] | [] | [] |
proofwiki-23606 | Rule of Commutation/Conjunction/Formulation 2/Proof 1/Forward Implication | :$\vdash \paren {p \land q} \implies \paren {q \land p}$ | {{BeginTableau|\vdash \paren {p \land q} \implies \paren {q \land p} }}
{{Assumption |1|\paren {p \land q} }}
{{Simplification|2|1|p|1|1}}
{{Simplification|3|1|q|1|2}}
{{Conjunction|4|1|\paren {q \land p}|3|2}}
{{Implication|5||\paren {p \land q} \implies \paren {q \land p}|1|4}}
{{EndTableau|lemma}} | :$\vdash \paren {p \land q} \implies \paren {q \land p}$ | {{BeginTableau|\vdash \paren {p \land q} \implies \paren {q \land p} }}
{{Assumption |1|\paren {p \land q} }}
{{Simplification|2|1|p|1|1}}
{{Simplification|3|1|q|1|2}}
{{Conjunction|4|1|\paren {q \land p}|3|2}}
{{Implication|5||\paren {p \land q} \implies \paren {q \land p}|1|4}}
{{EndTableau|lemma}} | Rule of Commutation/Conjunction/Formulation 2/Proof 1/Forward Implication | https://proofwiki.org/wiki/Rule_of_Commutation/Conjunction/Formulation_2/Proof_1/Forward_Implication | https://proofwiki.org/wiki/Rule_of_Commutation/Conjunction/Formulation_2/Proof_1/Forward_Implication | [
"Rule of Association"
] | [] | [] |
proofwiki-23607 | Rule of Commutation/Conjunction/Formulation 2/Proof 1/Reverse Implication | :$\vdash \paren {q \land p} \implies \paren {p \land q}$ | {{BeginTableau|\vdash \paren {q \land p} \implies \paren {p \land q} }}
{{Assumption |1|\paren {q \land p} }}
{{Simplification|2|1|q|1|1}}
{{Simplification|3|1|p|1|2}}
{{Conjunction|4|1|\paren {p \land q}|3|2}}
{{Implication|5||\paren {q \land p} \implies \paren {p \land q}|1|4}}
{{EndTableau|lemma}} | :$\vdash \paren {q \land p} \implies \paren {p \land q}$ | {{BeginTableau|\vdash \paren {q \land p} \implies \paren {p \land q} }}
{{Assumption |1|\paren {q \land p} }}
{{Simplification|2|1|q|1|1}}
{{Simplification|3|1|p|1|2}}
{{Conjunction|4|1|\paren {p \land q}|3|2}}
{{Implication|5||\paren {q \land p} \implies \paren {p \land q}|1|4}}
{{EndTableau|lemma}} | Rule of Commutation/Conjunction/Formulation 2/Proof 1/Reverse Implication | https://proofwiki.org/wiki/Rule_of_Commutation/Conjunction/Formulation_2/Proof_1/Reverse_Implication | https://proofwiki.org/wiki/Rule_of_Commutation/Conjunction/Formulation_2/Proof_1/Reverse_Implication | [
"Rule of Association"
] | [] | [] |
proofwiki-23608 | Characteristic Polynomial of Matrix is Monic | Let $R$ be a commutative ring with unity.
Let $\mathbf A$ be a square matrix over $R$ of order $n > 0$.
Let $\mathbf I_n$ be the $n \times n$ identity matrix.
Let $R \sqbrk x$ be the polynomial ring in one variable over $R$.
Let $\map {p_{\mathbf A} } x$ be the characteristic polynomial of $\mathbf A$.
Then $\map {p_{\... | From Existence of Schur Decomposition for Square Matrix, matrix $\mathbf A$ is similar to an upper triangular matrix $\mathbf T$.
Hence:
{{begin-eqn}}
{{eqn | l = \map {p_{\mathbf A} } x
| r = \map \det {x \mathbf I_n - \mathbf A}
| c = {{Defof|Characteristic Polynomial of Matrix}}
}}
{{eqn | r = \map \det ... | Let $R$ be a [[Definition:Commutative Ring with Unity|commutative ring with unity]].
Let $\mathbf A$ be a [[Definition:Square Matrix|square matrix]] over $R$ of [[Definition:Order of Square Matrix|order]] $n > 0$.
Let $\mathbf I_n$ be the $n \times n$ [[Definition:Identity Matrix|identity matrix]].
Let $R \sqbrk x$ ... | From [[Existence of Schur Decomposition for Square Matrix]], [[Definition:Matrix|matrix]] $\mathbf A$ is [[Definition:Matrix Similarity|similar]] to an [[Definition:Upper Triangular Matrix|upper triangular matrix]] $\mathbf T$.
Hence:
{{begin-eqn}}
{{eqn | l = \map {p_{\mathbf A} } x
| r = \map \det {x \mathbf I... | Characteristic Polynomial of Matrix is Monic | https://proofwiki.org/wiki/Characteristic_Polynomial_of_Matrix_is_Monic | https://proofwiki.org/wiki/Characteristic_Polynomial_of_Matrix_is_Monic | [
"Characteristic Polynomial of Matrix"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Matrix/Square Matrix",
"Definition:Matrix/Square Matrix/Order",
"Definition:Unit Matrix",
"Definition:Polynomial Ring",
"Definition:Characteristic Polynomial of Matrix",
"Definition:Monic Polynomial"
] | [
"Existence of Schur Decomposition for Square Matrix",
"Definition:Matrix",
"Definition:Matrix Similarity",
"Definition:Triangular Matrix/Upper Triangular Matrix",
"Similar Matrices have Same Characteristic Polynomial",
"Determinant of Upper Triangular Matrix",
"Definition:Main Diagonal/Diagonal Elements... |
proofwiki-23609 | Characteristic Polynomial of Triangular Matrix | Let $\mathbf T$ be a square triangular matrix of order $n$.
Then, the characteristic polynomial $\map {p_T} x$ is given by:
:$\ds \map {p_T} x = \prod_{i \mathop = 1}^n \paren {x - t_{ii} }$
where $t_{ii}$ are the diagonal elements of $\mathbf T$. | We have:
{{begin-eqn}}
{{eqn | l = \map {p_T} x
| r = \map \det {x \mathbf I - \mathbf T}
| c = {{Defof|Characteristic Polynomial of Matrix}}
}}
{{eqn | r = \prod_{i \mathop = 1}^n \paren {x - t_{ii} }
| c = Determinant of Triangular Matrix
}}
{{end-eqn}}
{{qed}}
Category:Triangular Matrices
Category:... | Let $\mathbf T$ be a [[Definition:Square Matrix|square]] [[Definition:Triangular Matrix|triangular matrix]] of order $n$.
Then, the [[Definition:Characteristic Polynomial of Matrix|characteristic polynomial]] $\map {p_T} x$ is given by:
:$\ds \map {p_T} x = \prod_{i \mathop = 1}^n \paren {x - t_{ii} }$
where $t_{ii}$... | We have:
{{begin-eqn}}
{{eqn | l = \map {p_T} x
| r = \map \det {x \mathbf I - \mathbf T}
| c = {{Defof|Characteristic Polynomial of Matrix}}
}}
{{eqn | r = \prod_{i \mathop = 1}^n \paren {x - t_{ii} }
| c = [[Determinant of Triangular Matrix]]
}}
{{end-eqn}}
{{qed}}
[[Category:Triangular Matrices]]... | Characteristic Polynomial of Triangular Matrix | https://proofwiki.org/wiki/Characteristic_Polynomial_of_Triangular_Matrix | https://proofwiki.org/wiki/Characteristic_Polynomial_of_Triangular_Matrix | [
"Triangular Matrices",
"Characteristic Polynomial of Matrix"
] | [
"Definition:Matrix/Square Matrix",
"Definition:Triangular Matrix",
"Definition:Characteristic Polynomial of Matrix",
"Definition:Main Diagonal/Diagonal Elements"
] | [
"Determinant of Triangular Matrix",
"Category:Triangular Matrices",
"Category:Characteristic Polynomial of Matrix"
] |
proofwiki-23610 | Eigenvalues of Triangular Matrix | The eigenvalues of a square triangular matrix are its diagonal elements. | Let $\mathbf T$ be a square triangular matrix.
From Characteristic Polynomial of Triangular Matrix, we have:
:$\ds \map {p_T} x = \prod_{i \mathop = 1}^n \paren {x - t_{ii} }$
where $t_{ii}$ are the diagonal elements of $\mathbf T$.
The eigenvalues are the roots of the characteristic polynomial of $\mathbf T$.
Thus, th... | The [[Definition:Eigenvalue of Square Matrix|eigenvalues]] of a [[Definition:Square Matrix|square]] [[Definition:Triangular Matrix|triangular matrix]] are its [[Definition:Diagonal Element|diagonal elements]]. | Let $\mathbf T$ be a [[Definition:Square Matrix|square]] [[Definition:Triangular Matrix|triangular matrix]].
From [[Characteristic Polynomial of Triangular Matrix]], we have:
:$\ds \map {p_T} x = \prod_{i \mathop = 1}^n \paren {x - t_{ii} }$
where $t_{ii}$ are the [[Definition:Diagonal Element|diagonal elements]] of $... | Eigenvalues of Triangular Matrix | https://proofwiki.org/wiki/Eigenvalues_of_Triangular_Matrix | https://proofwiki.org/wiki/Eigenvalues_of_Triangular_Matrix | [
"Triangular Matrices",
"Eigenvalues of Square Matrices"
] | [
"Definition:Eigenvalue/Square Matrix",
"Definition:Matrix/Square Matrix",
"Definition:Triangular Matrix",
"Definition:Main Diagonal/Diagonal Elements"
] | [
"Definition:Matrix/Square Matrix",
"Definition:Triangular Matrix",
"Characteristic Polynomial of Triangular Matrix",
"Definition:Main Diagonal/Diagonal Elements",
"Definition:Eigenvalue/Square Matrix",
"Definition:Root of Polynomial",
"Definition:Characteristic Polynomial of Matrix",
"Definition:Eigen... |
proofwiki-23611 | Trace of Matrix is Sum of Eigenvalues | Let $\mathbf A$ be a matrix of order $n$ over $\mathbb C$.
Let $\lambda_1, \lambda_2, \ldots, \lambda_n$ be the eigenvalues of $\mathbf A$ including algebraic multiplicity.
Then:
:$\ds \map \tr {\mathbf A} = \sum_{i \mathop = 1}^n \lambda_i$
where $\map \tr {\mathbf A}$ is the trace of $\mathbf A$. | From Existence of Schur Decomposition for Square Matrix, $\mathbf A$ is similar to an upper triangular matrix $\mathbf T$.
Let $\omega_1, \omega_2, \ldots, \omega_n$ be the eigenvalues of $\mathbf T$.
Therefore:
{{begin-eqn}}
{{eqn | l = \map \tr {\mathbf A}
| r = \map \tr {\mathbf T}
| c = Similar Matrices... | Let $\mathbf A$ be a [[Definition:Matrix|matrix]] of [[Definition:Order of Square Matrix|order]] $n$ over $\mathbb C$.
Let $\lambda_1, \lambda_2, \ldots, \lambda_n$ be the [[Definition:Eigenvalue of Square Matrix|eigenvalues]] of $\mathbf A$ including [[Definition:Algebraic Multiplicity|algebraic multiplicity]].
The... | From [[Existence of Schur Decomposition for Square Matrix]], $\mathbf A$ is [[Definition:Matrix Similarity|similar]] to an [[Definition:Upper Triangular Matrix|upper triangular matrix]] $\mathbf T$.
Let $\omega_1, \omega_2, \ldots, \omega_n$ be the [[Definition:Eigenvalue of Square Matrix|eigenvalues]] of $\mathbf T$.... | Trace of Matrix is Sum of Eigenvalues | https://proofwiki.org/wiki/Trace_of_Matrix_is_Sum_of_Eigenvalues | https://proofwiki.org/wiki/Trace_of_Matrix_is_Sum_of_Eigenvalues | [
"Traces of Matrices",
"Eigenvalues of Square Matrices"
] | [
"Definition:Matrix",
"Definition:Matrix/Square Matrix/Order",
"Definition:Eigenvalue/Square Matrix",
"Definition:Algebraic Multiplicity",
"Definition:Trace (Linear Algebra)/Matrix"
] | [
"Existence of Schur Decomposition for Square Matrix",
"Definition:Matrix Similarity",
"Definition:Triangular Matrix/Upper Triangular Matrix",
"Definition:Eigenvalue/Square Matrix",
"Similar Matrices have same Traces",
"Eigenvalues of Triangular Matrix",
"Similar Matrices have Same Eigenvalues"
] |
proofwiki-23612 | Hermitian Conjugate of Normal Matrix is Normal | The Hermitian conjugate of a normal matrix is normal. | Let $\mathbf N$ be a normal matrix.
We have:
{{begin-eqn}}
{{eqn | l = \mathbf N \mathbf N^\dagger
| r = \mathbf N^\dagger \mathbf N
| c = {{Defof|Normal Matrix}}
}}
{{eqn | l = \paren {\mathbf N^\dagger}^\dagger \paren {\mathbf N^\dagger}
| r = \paren {\mathbf N^\dagger} \paren {\mathbf N^\dagger}^\d... | The [[Definition:Hermitian Conjugate|Hermitian conjugate]] of a [[Definition:Normal Matrix|normal matrix]] is [[Definition:Normal Matrix|normal]]. | Let $\mathbf N$ be a [[Definition:Normal Matrix|normal matrix]].
We have:
{{begin-eqn}}
{{eqn | l = \mathbf N \mathbf N^\dagger
| r = \mathbf N^\dagger \mathbf N
| c = {{Defof|Normal Matrix}}
}}
{{eqn | l = \paren {\mathbf N^\dagger}^\dagger \paren {\mathbf N^\dagger}
| r = \paren {\mathbf N^\dagger}... | Hermitian Conjugate of Normal Matrix is Normal | https://proofwiki.org/wiki/Hermitian_Conjugate_of_Normal_Matrix_is_Normal | https://proofwiki.org/wiki/Hermitian_Conjugate_of_Normal_Matrix_is_Normal | [
"Normal Matrices",
"Hermitian Conjugates"
] | [
"Definition:Hermitian Conjugate",
"Definition:Normal Matrix",
"Definition:Normal Matrix"
] | [
"Definition:Normal Matrix",
"Hermitian Conjugate is Involution",
"Definition:Normal Matrix",
"Category:Normal Matrices",
"Category:Hermitian Conjugates"
] |
proofwiki-23613 | Lower Triangular Normal Matrix is Diagonal | A square matrix that is both lower triangular and normal is diagonal. | Let $\mathbf L$ be a square matrix that is both lower triangular and normal.
From Transpose of Upper Triangular Matrix is Lower Triangular, $\mathbf L^\intercal$ is upper triangular.
Since $\bar 0 = 0$, $\overline {\mathbf L^\intercal} = \mathbf L^\dagger$ is upper triangular.
From Hermitian Conjugate of Normal Matrix ... | A [[Definition:Square Matrix|square matrix]] that is both [[Definition:Lower Triangular Matrix|lower triangular]] and [[Definition:Normal Matrix|normal]] is [[Definition:Diagonal Matrix|diagonal]]. | Let $\mathbf L$ be a [[Definition:Square Matrix|square matrix]] that is both [[Definition:Lower Triangular Matrix|lower triangular]] and [[Definition:Normal Matrix|normal]].
From [[Transpose of Upper Triangular Matrix is Lower Triangular]], $\mathbf L^\intercal$ is [[Definition:Upper Triangular Matrix|upper triangular... | Lower Triangular Normal Matrix is Diagonal | https://proofwiki.org/wiki/Lower_Triangular_Normal_Matrix_is_Diagonal | https://proofwiki.org/wiki/Lower_Triangular_Normal_Matrix_is_Diagonal | [
"Triangular Normal Matrix is Diagonal",
"Normal Matrices",
"Lower Triangular Matrices",
"Diagonal Matrices"
] | [
"Definition:Matrix/Square Matrix",
"Definition:Triangular Matrix/Lower Triangular Matrix",
"Definition:Normal Matrix",
"Definition:Diagonal Matrix"
] | [
"Definition:Matrix/Square Matrix",
"Definition:Triangular Matrix/Lower Triangular Matrix",
"Definition:Normal Matrix",
"Transpose of Upper Triangular Matrix is Lower Triangular",
"Definition:Triangular Matrix/Upper Triangular Matrix",
"Definition:Triangular Matrix/Upper Triangular Matrix",
"Hermitian Co... |
proofwiki-23614 | Triangular Normal Matrix is Diagonal | A square matrix that is both triangular and normal is diagonal. | === Upper Triangular Normal Matrix is Diagonal ===
{{:Upper Triangular Normal Matrix is Diagonal}}{{qed|lemma}} | A [[Definition:Square Matrix|square matrix]] that is both [[Definition:Triangular Matrix|triangular]] and [[Definition:Normal Matrix|normal]] is [[Definition:Diagonal Matrix|diagonal]]. | === [[Upper Triangular Normal Matrix is Diagonal]] ===
{{:Upper Triangular Normal Matrix is Diagonal}}{{qed|lemma}} | Triangular Normal Matrix is Diagonal | https://proofwiki.org/wiki/Triangular_Normal_Matrix_is_Diagonal | https://proofwiki.org/wiki/Triangular_Normal_Matrix_is_Diagonal | [
"Triangular Normal Matrix is Diagonal",
"Triangular Matrices",
"Normal Matrices",
"Diagonal Matrices"
] | [
"Definition:Matrix/Square Matrix",
"Definition:Triangular Matrix",
"Definition:Normal Matrix",
"Definition:Diagonal Matrix"
] | [
"Upper Triangular Normal Matrix is Diagonal"
] |
proofwiki-23615 | (p implies q) implies q, q implies p therefore p/Theorem Form | :$\paren {\paren {\paren {p \implies q} \implies q} \land \paren {q \implies p} } \implies p$ | {{BeginTableau|\paren {\paren {\paren {p \implies q} \implies q} \land \paren {q \implies p} } \implies p}}
{{Assumption|1|\paren {\paren {p \implies q} \implies q} \land \paren {q \implies p} }}
{{TheoremIntro|2|\paren {\paren {p \implies q} \land \paren {q \implies r} } \implies \paren {p \implies r}|Hypothetical Syl... | :$\paren {\paren {\paren {p \implies q} \implies q} \land \paren {q \implies p} } \implies p$ | {{BeginTableau|\paren {\paren {\paren {p \implies q} \implies q} \land \paren {q \implies p} } \implies p}}
{{Assumption|1|\paren {\paren {p \implies q} \implies q} \land \paren {q \implies p} }}
{{TheoremIntro|2|\paren {\paren {p \implies q} \land \paren {q \implies r} } \implies \paren {p \implies r}|[[Hypothetical S... | (p implies q) implies q, q implies p therefore p/Theorem Form | https://proofwiki.org/wiki/(p_implies_q)_implies_q,_q_implies_p_therefore_p/Theorem_Form | https://proofwiki.org/wiki/(p_implies_q)_implies_q,_q_implies_p_therefore_p/Theorem_Form | [
"(p implies q) implies q, q implies p therefore p"
] | [] | [
"Hypothetical Syllogism/Formulation 3",
"Peirce's Law/Formulation 2"
] |
proofwiki-23616 | Finite Infimum in Subframe Equals Infimum in Frame | Let $L = \struct{S, \preceq}$ be a frame.
Let $T \subseteq S$.
Let $M = \struct{T, \preceq \restriction_T}$ be a subframe of $L$ where $\preceq \restriction_T$ is the restriction of $\preceq$ to $T$.
Let $F \subseteq T$ be finite.
The infimum ${\bigwedge}_M F$ of $F$ exists in $M$ and equals the infimum $\bigvee F$ in ... | {{Recall|Subframe|subframe}}
{{:Definition:Subframe}}
From Infimum in Ordered Subset:
:$\forall $ finite $F \subseteq T$, the infimum ${\bigwedge}_M F$ of $F$ exists in $M$ and ${\bigvee}_M F {{=}} \bigvee F$
{{qed}}
Category:Subframes
Category:Frames
fm06rx5ji40p569es64ppu97uj9312n | Let $L = \struct{S, \preceq}$ be a [[Definition:Frame (Lattice Theory)|frame]].
Let $T \subseteq S$.
Let $M = \struct{T, \preceq \restriction_T}$ be a [[Definition:Subframe|subframe]] of $L$ where $\preceq \restriction_T$ is the [[Definition:Restriction of Ordering|restriction]] of $\preceq$ to $T$.
Let $F \subset... | {{Recall|Subframe|subframe}}
{{:Definition:Subframe}}
From [[Infimum in Ordered Subset]]:
:$\forall $ [[Definition:Finite Subset|finite]] $F \subseteq T$, the [[Definition:Infimum of Set|infimum]] ${\bigwedge}_M F$ of $F$ exists in $M$ and ${\bigvee}_M F {{=}} \bigvee F$
{{qed}}
[[Category:Subframes]]
[[Category:Fram... | Finite Infimum in Subframe Equals Infimum in Frame | https://proofwiki.org/wiki/Finite_Infimum_in_Subframe_Equals_Infimum_in_Frame | https://proofwiki.org/wiki/Finite_Infimum_in_Subframe_Equals_Infimum_in_Frame | [
"Subframes",
"Frames"
] | [
"Definition:Frame (Lattice Theory)",
"Definition:Subframe",
"Definition:Restriction of Ordering",
"Definition:Finite Subset",
"Definition:Infimum of Set",
"Definition:Infimum of Set"
] | [
"Infimum in Ordered Subset",
"Definition:Finite Subset",
"Definition:Infimum of Set",
"Category:Subframes",
"Category:Frames"
] |
proofwiki-23617 | Increasing Sum of Squares of Binomial Coefficients | :$\ds \sum_{r \mathop = 1}^n r \binom n r^2 = \dfrac {\paren {2 n - 1}!} {\paren {\paren {n - 1}!}^2}$
where $\dbinom n r$ denotes the binomial coefficient:
:$\dbinom n r = \dfrac {n!} {r! \paren {n - r} }$ | We note that $\ds \sum_{r = 1}^n r \binom n r^2$ is the same as $\ds \sum_{r = 0}^n r \binom n r^2$ because the zeroth term is $0 \dbinom n 0^2 = 0$.
Then:
{{begin-eqn}}
{{eqn | n = 1
| l = \sum_{r \mathop = 0}^n r \binom n r^2
| r = 0 \binom n 0^2 + 1 \binom n 1^2 + 2 \binom n 2^2 + 3 \binom n 3^2 + \cdots... | :$\ds \sum_{r \mathop = 1}^n r \binom n r^2 = \dfrac {\paren {2 n - 1}!} {\paren {\paren {n - 1}!}^2}$
where $\dbinom n r$ denotes the [[Definition:Binomial Coefficient|binomial coefficient]]:
:$\dbinom n r = \dfrac {n!} {r! \paren {n - r} }$ | We note that $\ds \sum_{r = 1}^n r \binom n r^2$ is the same as $\ds \sum_{r = 0}^n r \binom n r^2$ because the zeroth term is $0 \dbinom n 0^2 = 0$.
Then:
{{begin-eqn}}
{{eqn | n = 1
| l = \sum_{r \mathop = 0}^n r \binom n r^2
| r = 0 \binom n 0^2 + 1 \binom n 1^2 + 2 \binom n 2^2 + 3 \binom n 3^2 + \cdo... | Increasing Sum of Squares of Binomial Coefficients | https://proofwiki.org/wiki/Increasing_Sum_of_Squares_of_Binomial_Coefficients | https://proofwiki.org/wiki/Increasing_Sum_of_Squares_of_Binomial_Coefficients | [
"Binomial Coefficients"
] | [
"Definition:Binomial Coefficient"
] | [
"Symmetry Rule for Binomial Coefficients",
"Sum of Squares of Binomial Coefficients",
"Definition:Fraction/Numerator",
"Definition:Fraction/Denominator"
] |
proofwiki-23618 | Image of Frame under Frame Homomorphism is Frame | Let $L_1 = \struct {S_1, \preceq_1}$ and $L_2 = \struct {S_2, \preceq_2}$ be frames.
Let $\phi: S_1 \to S_2$ be a frame homomorphism.
Let $L_3 = \struct{S_3, \preceq_3}$ be the ordered set where:
:$S_3$ denotes the image of $S_1$ under $\phi$
:$\preceq_3$ denotes the restricted ordering of $\preceq_2$ to the subset $S_... | === {{Lemma|Image of Frame under Frame Homomorphism is Frame|1|Lemma 1}} ===
{{:Image of Frame under Frame Homomorphism is Frame/Lemma 1}}{{qed|lemma}} | Let $L_1 = \struct {S_1, \preceq_1}$ and $L_2 = \struct {S_2, \preceq_2}$ be [[Definition:Frame (Lattice Theory)|frames]].
Let $\phi: S_1 \to S_2$ be a [[Definition:Frame Homomorphism|frame homomorphism]].
Let $L_3 = \struct{S_3, \preceq_3}$ be the [[Definition:Ordered Set|ordered set]] where:
:$S_3$ denotes the [[D... | === {{Lemma|Image of Frame under Frame Homomorphism is Frame|1|Lemma 1}} ===
{{:Image of Frame under Frame Homomorphism is Frame/Lemma 1}}{{qed|lemma}} | Image of Frame under Frame Homomorphism is Frame | https://proofwiki.org/wiki/Image_of_Frame_under_Frame_Homomorphism_is_Frame | https://proofwiki.org/wiki/Image_of_Frame_under_Frame_Homomorphism_is_Frame | [
"Image of Frame under Frame Homomorphism is Frame",
"Frame Homomorphisms",
"Frames"
] | [
"Definition:Frame (Lattice Theory)",
"Definition:Frame Homomorphism",
"Definition:Ordered Set",
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Restriction of Ordering",
"Definition:Subset",
"Definition:Subframe"
] | [] |
proofwiki-23619 | Image of Frame under Frame Homomorphism is Frame/Lemma 1 | Let $T \subseteq S_3$.
Let $\bigvee T$ denote the supremum of $T$ in $L_2$.
Then:
:$\bigvee T \in S_3$. | {{Recall|Frame (Lattice Theory)|frame}}
{{:Definition:Frame (Lattice Theory)}}
{{Recall|Complete Lattice|complete Lattice}}
{{:Definition:Complete Lattice/Definition 1}}
By definition of image:
:$S_3 = \phi \sqbrk {S_1}$
From {{Corollary|Image of Preimage under Mapping}}:
:$(1) \quad T = \phi \sqbrk {\phi^{-1} \sqbrk T... | Let $T \subseteq S_3$.
Let $\bigvee T$ denote the [[Definition:Supremum of Set|supremum]] of $T$ in $L_2$.
Then:
:$\bigvee T \in S_3$. | {{Recall|Frame (Lattice Theory)|frame}}
{{:Definition:Frame (Lattice Theory)}}
{{Recall|Complete Lattice|complete Lattice}}
{{:Definition:Complete Lattice/Definition 1}}
By definition of [[Definition:Image of Subset under Mapping|image]]:
:$S_3 = \phi \sqbrk {S_1}$
From {{Corollary|Image of Preimage under Mapping}}:... | Image of Frame under Frame Homomorphism is Frame/Lemma 1 | https://proofwiki.org/wiki/Image_of_Frame_under_Frame_Homomorphism_is_Frame/Lemma_1 | https://proofwiki.org/wiki/Image_of_Frame_under_Frame_Homomorphism_is_Frame/Lemma_1 | [
"Image of Frame under Frame Homomorphism is Frame"
] | [
"Definition:Supremum of Set"
] | [
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Image (Set Theory)/Mapping/Subset",
"Category:Image of Frame under Frame Homomorphism is Frame"
] |
proofwiki-23620 | Image of Frame under Frame Homomorphism is Frame/Lemma 2 | Let $F \subseteq S_3$ be a finite subset.
Let $\bigwedge F$ denote the infimum of $F$ in $L_2$.
Then:
:$\bigwedge F \in S_3$. | {{Recall|Frame (Lattice Theory)|frame}}
{{:Definition:Frame (Lattice Theory)}}
{{Recall|Complete Lattice|complete Lattice}}
{{:Definition:Complete Lattice/Definition 1}}
By definition of image:
:$\forall y \in F : \exists x \in S_1 : \map \phi x = y$
For each $y \in F$ choose $x_y \in S_1 : \map \phi {x_y} = y$.
Let $G... | Let $F \subseteq S_3$ be a [[Definition:Finite Subset|finite subset]].
Let $\bigwedge F$ denote the [[Definition:Infimum of Set|infimum]] of $F$ in $L_2$.
Then:
:$\bigwedge F \in S_3$. | {{Recall|Frame (Lattice Theory)|frame}}
{{:Definition:Frame (Lattice Theory)}}
{{Recall|Complete Lattice|complete Lattice}}
{{:Definition:Complete Lattice/Definition 1}}
By definition of [[Definition:Image of Subset under Mapping|image]]:
:$\forall y \in F : \exists x \in S_1 : \map \phi x = y$
For each $y \in F$ c... | Image of Frame under Frame Homomorphism is Frame/Lemma 2 | https://proofwiki.org/wiki/Image_of_Frame_under_Frame_Homomorphism_is_Frame/Lemma_2 | https://proofwiki.org/wiki/Image_of_Frame_under_Frame_Homomorphism_is_Frame/Lemma_2 | [
"Image of Frame under Frame Homomorphism is Frame"
] | [
"Definition:Finite Subset",
"Definition:Infimum of Set"
] | [
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Image (Set Theory)/Mapping/Subset",
"Category:Image of Frame under Frame Homomorphism is Frame"
] |
proofwiki-23621 | Normal Matrix has Orthogonal Eigenspaces | Let $\mathbf N$ be a normal matrix.
Let $\lambda_1, \lambda_2$ be distinct eigenvalues of $\mathbf N$.
Then, the eigenspaces of $\lambda_1$ and $\lambda_2$ are orthogonal:
:$\map \ker {\mathbf N - \lambda_1 \mathbf I} \perp \map \ker {\mathbf N - \lambda_2 \mathbf I}$
where:
:$\ker$ denotes kernel
:$\mathbf I$ denotes ... | Follows immediately from Normal Matrix is Normal Operator and Eigenvalues of Normal Operator have Orthogonal Eigenspaces
{{qed}}
Category:Normal Matrices
Category:Eigenspaces
fte1v1yw5dz4nzxkd9c2l4vks7vbz81 | Let $\mathbf N$ be a [[Definition:Normal Matrix|normal matrix]].
Let $\lambda_1, \lambda_2$ be distinct [[Definition:Eigenvalue of Linear Operator|eigenvalues]] of $\mathbf N$.
Then, the [[Definition:Eigenspace of Linear Operator|eigenspaces]] of $\lambda_1$ and $\lambda_2$ are [[Definition:Orthogonal Sets|orthogona... | Follows immediately from [[Normal Matrix is Normal Operator]] and [[Eigenvalues of Normal Operator have Orthogonal Eigenspaces]]
{{qed}}
[[Category:Normal Matrices]]
[[Category:Eigenspaces]]
fte1v1yw5dz4nzxkd9c2l4vks7vbz81 | Normal Matrix has Orthogonal Eigenspaces | https://proofwiki.org/wiki/Normal_Matrix_has_Orthogonal_Eigenspaces | https://proofwiki.org/wiki/Normal_Matrix_has_Orthogonal_Eigenspaces | [
"Normal Matrices",
"Eigenspaces"
] | [
"Definition:Normal Matrix",
"Definition:Eigenvalue/Linear Operator",
"Definition:Eigenspace/Linear Operator",
"Definition:Orthogonal (Linear Algebra)/Sets",
"Definition:Kernel of Linear Transformation",
"Definition:Unit Matrix",
"Definition:Orthogonal (Linear Algebra)/Sets"
] | [
"Normal Matrix is Normal Operator",
"Eigenvalues of Normal Operator have Orthogonal Eigenspaces",
"Category:Normal Matrices",
"Category:Eigenspaces"
] |
proofwiki-23622 | Inclusion Mapping of Subframe into Frame is Frame Homomorphism | Let $L = \struct {S, \preceq}$ be a frame.
Let $T \subseteq S$.
Let $M = \struct {T, \preceq \restriction_T}$ be a subframe of $L$ where $\preceq \restriction_T$ is the restriction of $\preceq$ to $T$.
The inclusion mapping $i: M \to L$ is a frame homomorphism. | {{Recall|theorem = Supremum in Subframe Equals Supremum in Frame}}
{{:Supremum in Subframe Equals Supremum in Frame}}
{{Recall|theorem = Finite Infimum in Subframe Equals Infimum in Frame}}
{{:Finite Infimum in Subframe Equals Infimum in Frame}}
We have:
{{begin-eqn}}
{{eqn | q = \forall A \subseteq T
| l = \map ... | Let $L = \struct {S, \preceq}$ be a [[Definition:Frame (Lattice Theory)|frame]].
Let $T \subseteq S$.
Let $M = \struct {T, \preceq \restriction_T}$ be a [[Definition:Subframe|subframe]] of $L$ where $\preceq \restriction_T$ is the [[Definition:Restriction of Ordering|restriction]] of $\preceq$ to $T$.
The [[Defini... | {{Recall|theorem = Supremum in Subframe Equals Supremum in Frame}}
{{:Supremum in Subframe Equals Supremum in Frame}}
{{Recall|theorem = Finite Infimum in Subframe Equals Infimum in Frame}}
{{:Finite Infimum in Subframe Equals Infimum in Frame}}
We have:
{{begin-eqn}}
{{eqn | q = \forall A \subseteq T
| l = \m... | Inclusion Mapping of Subframe into Frame is Frame Homomorphism | https://proofwiki.org/wiki/Inclusion_Mapping_of_Subframe_into_Frame_is_Frame_Homomorphism | https://proofwiki.org/wiki/Inclusion_Mapping_of_Subframe_into_Frame_is_Frame_Homomorphism | [
"Subframes",
"Frame Homomorphisms",
"Inclusion Mappings"
] | [
"Definition:Frame (Lattice Theory)",
"Definition:Subframe",
"Definition:Restriction of Ordering",
"Definition:Inclusion Mapping",
"Definition:Frame Homomorphism"
] | [
"Category:Subframes",
"Category:Frame Homomorphisms",
"Category:Inclusion Mappings"
] |
proofwiki-23623 | Supremum in Subframe Equals Supremum in Frame | Let $L = \struct{S, \preceq}$ be a frame.
Let $T \subseteq S$.
Let $M = \struct{T, \preceq \restriction_T}$ be a subframe of $L$ where $\preceq \restriction_T$ is the restriction of $\preceq$ to $T$.
Let $A \subseteq T$.
The supremum ${\bigvee}_M A$ of $A$ exists in $M$ and equals the supremum $\bigvee A$ in $L$. | {{Recall|Subframe|subframe}}
{{:Definition:Subframe}}
From Supremum in Ordered Subset:
:$\forall A \subseteq T$, the supremum ${\bigvee}_M A$ of $A$ exists in $M$ and ${\bigvee}_M A {{=}} \bigvee A$
{{qed}}
Category:Subframes
Category:Frames
p3ghvw0nb6ahqffgw770zo3k2htzyyh | Let $L = \struct{S, \preceq}$ be a [[Definition:Frame (Lattice Theory)|frame]].
Let $T \subseteq S$.
Let $M = \struct{T, \preceq \restriction_T}$ be a [[Definition:Subframe|subframe]] of $L$ where $\preceq \restriction_T$ is the [[Definition:Restriction of Ordering|restriction]] of $\preceq$ to $T$.
Let $A \subset... | {{Recall|Subframe|subframe}}
{{:Definition:Subframe}}
From [[Supremum in Ordered Subset]]:
:$\forall A \subseteq T$, the [[Definition:Supremum of Set|supremum]] ${\bigvee}_M A$ of $A$ exists in $M$ and ${\bigvee}_M A {{=}} \bigvee A$
{{qed}}
[[Category:Subframes]]
[[Category:Frames]]
p3ghvw0nb6ahqffgw770zo3k2htzyyh | Supremum in Subframe Equals Supremum in Frame | https://proofwiki.org/wiki/Supremum_in_Subframe_Equals_Supremum_in_Frame | https://proofwiki.org/wiki/Supremum_in_Subframe_Equals_Supremum_in_Frame | [
"Subframes",
"Frames"
] | [
"Definition:Frame (Lattice Theory)",
"Definition:Subframe",
"Definition:Restriction of Ordering",
"Definition:Supremum of Set",
"Definition:Supremum of Set"
] | [
"Supremum in Ordered Subset",
"Definition:Supremum of Set",
"Category:Subframes",
"Category:Frames"
] |
proofwiki-23624 | Subframe is a Frame | Let $L = \struct{S, \preceq}$ be a frame.
Let $T \subseteq S$.
Let $M = \struct{T, \preceq \restriction_T}$ be a subframe of $L$ where $\preceq \restriction_T$ is the restriction of $\preceq$ to $T$.
Then $M$ is a frame. | {{Recall|Frame (Lattice Theory)|frame}}
{{:Definition:Frame (Lattice Theory)}} | Let $L = \struct{S, \preceq}$ be a [[Definition:Frame (Lattice Theory)|frame]].
Let $T \subseteq S$.
Let $M = \struct{T, \preceq \restriction_T}$ be a [[Definition:Subframe|subframe]] of $L$ where $\preceq \restriction_T$ is the [[Definition:Restriction of Ordering|restriction]] of $\preceq$ to $T$.
Then $M$ is a ... | {{Recall|Frame (Lattice Theory)|frame}}
{{:Definition:Frame (Lattice Theory)}} | Subframe is a Frame | https://proofwiki.org/wiki/Subframe_is_a_Frame | https://proofwiki.org/wiki/Subframe_is_a_Frame | [
"Subframes",
"Frames"
] | [
"Definition:Frame (Lattice Theory)",
"Definition:Subframe",
"Definition:Restriction of Ordering",
"Definition:Frame (Lattice Theory)"
] | [] |
proofwiki-23625 | Greatest Term of Binomial Expansion/Examples/Arbitrary Example 2 | Consider the expression:
:$E = \paren {1 - 3 x}^{-\frac 7 3}$
Let $x = \dfrac 1 4$.
Then the greatest terms in the power series expansion of $E$ by means of the General Binomial Theorem are:
{{begin-eqn}}
{{eqn | o =
| r = \dfrac {7 \times 10 \times 13} {4^3 \times 3!}
}}
{{eqn | r = \size {-\dfrac {7 \times 10 ... | Let us perform the expansion:
{{begin-eqn}}
{{eqn | l = \paren {1 - 3 x}^{-\frac 7 3}
| r = 1 + \paren {-\dfrac 4 3 - 1} \paren {-3 x} + \dfrac {\paren {-\frac 4 3 - 1} \paren {-\frac 4 3 - 2} } {2!} \paren {-3 x}^2 + \dfrac {\paren {-\frac 4 3 - 1} \paren {-\frac 4 3 - 2} \paren {-\frac 4 3 - 3} } {3!} \paren {-... | Consider the [[Definition:Expression|expression]]:
:$E = \paren {1 - 3 x}^{-\frac 7 3}$
Let $x = \dfrac 1 4$.
Then the greatest terms in the [[Definition:Power Series Expansion|power series expansion]] of $E$ by means of the [[General Binomial Theorem]] are:
{{begin-eqn}}
{{eqn | o =
| r = \dfrac {7 \times ... | Let us perform the expansion:
{{begin-eqn}}
{{eqn | l = \paren {1 - 3 x}^{-\frac 7 3}
| r = 1 + \paren {-\dfrac 4 3 - 1} \paren {-3 x} + \dfrac {\paren {-\frac 4 3 - 1} \paren {-\frac 4 3 - 2} } {2!} \paren {-3 x}^2 + \dfrac {\paren {-\frac 4 3 - 1} \paren {-\frac 4 3 - 2} \paren {-\frac 4 3 - 3} } {3!} \paren {... | Greatest Term of Binomial Expansion/Examples/Arbitrary Example 2 | https://proofwiki.org/wiki/Greatest_Term_of_Binomial_Expansion/Examples/Arbitrary_Example_2 | https://proofwiki.org/wiki/Greatest_Term_of_Binomial_Expansion/Examples/Arbitrary_Example_2 | [
"Binomial Theorem"
] | [
"Definition:Expression",
"Definition:Power Series Expansion",
"Binomial Theorem/General Binomial Theorem"
] | [
"Definition:Sign of Number",
"Definition:Coefficient of Polynomial",
"Definition:Absolute Value"
] |
proofwiki-23626 | Simple Spectrum Approximation of Normal Matrix/Property 1 | :$\ds \lim_{t \mathop \to 0} \map {\mathbf A} t = \mathbf A$ | {{begin-eqn}}
{{eqn | l = \lim_{t \mathop \to 0} \map {\mathbf A} t
| r = \map {\mathbf A} 0
}}
{{eqn | r = \mathbf A + 0 \mathbf U \mathbf E \mathbf U^\dagger
}}
{{eqn | r = \mathbf A
}}
{{end-eqn}}
{{qed}}
Category:Simple Spectrum Approximation of Normal Matrix
9ikjtwsrk54q2ommh9xft6xaa5e41fr | :$\ds \lim_{t \mathop \to 0} \map {\mathbf A} t = \mathbf A$ | {{begin-eqn}}
{{eqn | l = \lim_{t \mathop \to 0} \map {\mathbf A} t
| r = \map {\mathbf A} 0
}}
{{eqn | r = \mathbf A + 0 \mathbf U \mathbf E \mathbf U^\dagger
}}
{{eqn | r = \mathbf A
}}
{{end-eqn}}
{{qed}}
[[Category:Simple Spectrum Approximation of Normal Matrix]]
9ikjtwsrk54q2ommh9xft6xaa5e41fr | Simple Spectrum Approximation of Normal Matrix/Property 1 | https://proofwiki.org/wiki/Simple_Spectrum_Approximation_of_Normal_Matrix/Property_1 | https://proofwiki.org/wiki/Simple_Spectrum_Approximation_of_Normal_Matrix/Property_1 | [
"Simple Spectrum Approximation of Normal Matrix"
] | [] | [
"Category:Simple Spectrum Approximation of Normal Matrix"
] |
proofwiki-23627 | Simple Spectrum Approximation of Normal Matrix/Property 2 | :$\map {\mathbf A} t$ is normal for all values of $t$ | Let $\mathbf D = \mathbf U^\dagger \mathbf D \mathbf U$ be the diagonalized version of $\mathbf A$.
We have:
{{begin-eqn}}
{{eqn | l = \map {\mathbf A} t
| r = \mathbf A + t \mathbf U \mathbf E \mathbf U^\dagger
}}
{{eqn | r = \mathbf U \mathbf D \mathbf U^\dagger + t \mathbf U \mathbf E \mathbf U^\dagger
}}
{{eq... | :$\map {\mathbf A} t$ is [[Definition:Normal Matrix|normal]] for all values of $t$ | Let $\mathbf D = \mathbf U^\dagger \mathbf D \mathbf U$ be the [[Definition:Diagonalizable Matrix|diagonalized]] version of $\mathbf A$.
We have:
{{begin-eqn}}
{{eqn | l = \map {\mathbf A} t
| r = \mathbf A + t \mathbf U \mathbf E \mathbf U^\dagger
}}
{{eqn | r = \mathbf U \mathbf D \mathbf U^\dagger + t \mathbf... | Simple Spectrum Approximation of Normal Matrix/Property 2 | https://proofwiki.org/wiki/Simple_Spectrum_Approximation_of_Normal_Matrix/Property_2 | https://proofwiki.org/wiki/Simple_Spectrum_Approximation_of_Normal_Matrix/Property_2 | [
"Simple Spectrum Approximation of Normal Matrix"
] | [
"Definition:Normal Matrix"
] | [
"Definition:Diagonalizable Matrix",
"Definition:Diagonal Matrix",
"Definition:Diagonal Matrix",
"Spectral Theorem for Normal Matrices",
"Definition:Unitary Matrix",
"Definition:Matrix Similarity",
"Definition:Diagonal Matrix",
"Definition:Normal Matrix",
"Category:Simple Spectrum Approximation of No... |
proofwiki-23628 | Simple Spectrum Approximation of Normal Matrix/Property 3 | :$\map {\mathbf A} t$ has the same eigenvectors as $\mathbf A$ for all values of $t$ | From Property $(2)$ of Simple Spectrum Approximation of Normal Matrix, $\map {\mathbf A} t$ is normal.
From Spectral Theorem for Normal Matrices, $\map {\mathbf A} t$ is diagonalizable.
From Simple Spectrum Approximation of Diagonalizable Matrix, $\map {\mathbf A} t$ has the same eigenvectors as $\mathbf A$.
{{qed}}
Ca... | :$\map {\mathbf A} t$ has the same [[Definition:Eigenvector of Square Matrix|eigenvectors]] as $\mathbf A$ for all values of $t$ | From Property $(2)$ of [[Simple Spectrum Approximation of Normal Matrix]], $\map {\mathbf A} t$ is [[Definition:Normal Matrix|normal]].
From [[Spectral Theorem for Normal Matrices]], $\map {\mathbf A} t$ is [[Definition:Diagonalizable Matrix|diagonalizable]].
From [[Simple Spectrum Approximation of Diagonalizable Mat... | Simple Spectrum Approximation of Normal Matrix/Property 3 | https://proofwiki.org/wiki/Simple_Spectrum_Approximation_of_Normal_Matrix/Property_3 | https://proofwiki.org/wiki/Simple_Spectrum_Approximation_of_Normal_Matrix/Property_3 | [
"Simple Spectrum Approximation of Normal Matrix"
] | [
"Definition:Eigenvector/Square Matrix"
] | [
"Simple Spectrum Approximation of Normal Matrix",
"Definition:Normal Matrix",
"Spectral Theorem for Normal Matrices",
"Definition:Diagonalizable Matrix",
"Simple Spectrum Approximation of Diagonalizable Matrix",
"Definition:Eigenvector/Square Matrix",
"Category:Simple Spectrum Approximation of Normal Ma... |
proofwiki-23629 | Simple Spectrum Approximation of Normal Matrix/Property 5 | :There exists $\tau > 0$ such that $0 < t < \tau \implies \map \sigma {\map {\mathbf A} t}$ is simple. | From Property $(2)$ of Simple Spectrum Approximation of Normal Matrix, $\map {\mathbf A} t$ is normal.
From Spectral Theorem for Normal Matrices, $\map {\mathbf A} t$ is diagonalizable.
From Simple Spectrum Approximation of Diagonalizable Matrix, there exists $\tau > 0$ such that:
:$0 < t < \tau \implies \map \sigma {\... | :There exists $\tau > 0$ such that $0 < t < \tau \implies \map \sigma {\map {\mathbf A} t}$ is [[Definition:Simple Eigenvalue|simple]]. | From Property $(2)$ of [[Simple Spectrum Approximation of Normal Matrix]], $\map {\mathbf A} t$ is [[Definition:Normal Matrix|normal]].
From [[Spectral Theorem for Normal Matrices]], $\map {\mathbf A} t$ is [[Definition:Diagonalizable Matrix|diagonalizable]].
From [[Simple Spectrum Approximation of Diagonalizable Mat... | Simple Spectrum Approximation of Normal Matrix/Property 5 | https://proofwiki.org/wiki/Simple_Spectrum_Approximation_of_Normal_Matrix/Property_5 | https://proofwiki.org/wiki/Simple_Spectrum_Approximation_of_Normal_Matrix/Property_5 | [
"Simple Spectrum Approximation of Normal Matrix"
] | [
"Definition:Simple Eigenvalue"
] | [
"Simple Spectrum Approximation of Normal Matrix",
"Definition:Normal Matrix",
"Spectral Theorem for Normal Matrices",
"Definition:Diagonalizable Matrix",
"Simple Spectrum Approximation of Diagonalizable Matrix",
"Definition:Simple Eigenvalue",
"Category:Simple Spectrum Approximation of Normal Matrix"
] |
proofwiki-23630 | Halting Problem is Undecidable | No Turing machine can be programmed to determine whether an arbitrary Turing machine will halt on arbitrary input. | {{AimForCont}} there exists a Turing machine $\map H {T, I}$ that determines whether another Turing machine $T$ halts with input $I$:
:$\map H {T, I} = \begin{cases} \text{true} & : T \text{ halts on input } I \\ \text{false} & : T \text{ never halts on input } I \end{cases}$
Then, the following Turing machine can be i... | No [[Definition:Turing Machine|Turing machine]] can be programmed to determine whether an [[Definition:Arbitrary|arbitrary]] [[Definition:Turing Machine|Turing machine]] will [[Definition:Halt|halt]] on [[Definition:Arbitrary|arbitrary]] input. | {{AimForCont}} there exists a [[Definition:Turing Machine|Turing machine]] $\map H {T, I}$ that determines whether another [[Definition:Turing Machine|Turing machine]] $T$ [[Definition:Halt|halts]] with input $I$:
:$\map H {T, I} = \begin{cases} \text{true} & : T \text{ halts on input } I \\ \text{false} & : T \text{ n... | Halting Problem is Undecidable | https://proofwiki.org/wiki/Halting_Problem_is_Undecidable | https://proofwiki.org/wiki/Halting_Problem_is_Undecidable | [
"Halting Problem",
"Turing Machines"
] | [
"Definition:Turing Machine",
"Definition:Arbitrary",
"Definition:Turing Machine",
"Definition:Halt",
"Definition:Arbitrary"
] | [
"Definition:Turing Machine",
"Definition:Turing Machine",
"Definition:Halt",
"Definition:Turing Machine",
"Definition:Halt",
"Definition:Halt",
"Definition:Halt",
"Definition:Halt",
"Definition:Halt",
"Definition:Contradiction",
"Category:Halting Problem",
"Category:Turing Machines"
] |
proofwiki-23631 | Sum from 0 to Infinity of (r+1) x^r | Let $x \in \R$ such that $\size x < 1$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{r \mathop = 0}^\infty \paren {r + 1} x^r
| r = 1 + 2 x + 3 x^2 + 4 x^3 + \cdots
| c =
}}
{{eqn | r = \dfrac 1 {\paren {1 - x}^2}
| c =
}}
{{end-eqn}} | From Sum from $0$ to $n - 1$ of $\paren {r + 1} x^r$:
:$\ds \sum_{r \mathop = 0}^{n - 1} \paren {r + 1} x^r = \dfrac {1 - x^n} {\paren {1 - x}^2} - \dfrac {n x^n} {1 - x}$
From Real Null Sequence: $n^\alpha x^n$:
:$\ds \lim_{n \mathop \to \infty} n x^n = 0$
Then we have that:
:$\ds \lim_{n \mathop \to \infty} x^n = 0$
... | Let $x \in \R$ such that $\size x < 1$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{r \mathop = 0}^\infty \paren {r + 1} x^r
| r = 1 + 2 x + 3 x^2 + 4 x^3 + \cdots
| c =
}}
{{eqn | r = \dfrac 1 {\paren {1 - x}^2}
| c =
}}
{{end-eqn}} | From [[Sum from 0 to n-1 of (r+1) x^r|Sum from $0$ to $n - 1$ of $\paren {r + 1} x^r$]]:
:$\ds \sum_{r \mathop = 0}^{n - 1} \paren {r + 1} x^r = \dfrac {1 - x^n} {\paren {1 - x}^2} - \dfrac {n x^n} {1 - x}$
From [[Real Null Sequence/Examples/n^alpha x^n|Real Null Sequence: $n^\alpha x^n$]]:
:$\ds \lim_{n \mathop \to ... | Sum from 0 to Infinity of (r+1) x^r/Proof 1 | https://proofwiki.org/wiki/Sum_from_0_to_Infinity_of_(r+1)_x^r | https://proofwiki.org/wiki/Sum_from_0_to_Infinity_of_(r+1)_x^r/Proof_1 | [
"Sum from 0 to Infinity of (r+1) x^r",
"Examples of Power Series"
] | [] | [
"Sum from 0 to n-1 of (r+1) x^r",
"Real Null Sequence/Examples/n^alpha x^n"
] |
proofwiki-23632 | Dirac Bra Grouping is Conjugate Linear | Let $\bra \alpha$ and $\bra \beta$ be bras in some dual space.
Let $h$ and $k$ be scalars.
Then:
:$\bra {h \alpha + k \beta} = \bra \alpha h^\dagger + \bra \beta k^\dagger$
where $h^\dagger$ is the complex conjugate of $h$. | {{begin-eqn}}
{{eqn | l = \bra {h \alpha + k \beta}
| r = \ket {h \alpha + k \beta}^*
| c = {{Defof|Dirac Notation/Bra|Bra}}
}}
{{eqn | r = \paren {h \ket \alpha + k \ket \beta}^*
| c = {{Defof|Dirac Notation/Ket/Abuse of Notation|Ket notation}}
}}
{{eqn | r = h^\dagger \ket \alpha^* + k^\dagger \ket ... | Let $\bra \alpha$ and $\bra \beta$ be [[Definition:Dirac Notation/Bra|bras]] in some [[Definition:Dual Vector Space|dual space]].
Let $h$ and $k$ be [[Definition:Scalar|scalars]].
Then:
:$\bra {h \alpha + k \beta} = \bra \alpha h^\dagger + \bra \beta k^\dagger$
where $h^\dagger$ is the [[Definition:Complex Conjugate... | {{begin-eqn}}
{{eqn | l = \bra {h \alpha + k \beta}
| r = \ket {h \alpha + k \beta}^*
| c = {{Defof|Dirac Notation/Bra|Bra}}
}}
{{eqn | r = \paren {h \ket \alpha + k \ket \beta}^*
| c = {{Defof|Dirac Notation/Ket/Abuse of Notation|Ket notation}}
}}
{{eqn | r = h^\dagger \ket \alpha^* + k^\dagger \ket ... | Dirac Bra Grouping is Conjugate Linear | https://proofwiki.org/wiki/Dirac_Bra_Grouping_is_Conjugate_Linear | https://proofwiki.org/wiki/Dirac_Bra_Grouping_is_Conjugate_Linear | [
"Bras",
"Dirac Notation"
] | [
"Definition:Dirac Notation/Bra",
"Definition:Algebraic Dual/Vector Space",
"Definition:Scalar",
"Definition:Complex Conjugate"
] | [
"Adjoint is Conjugate Linear",
"Category:Bras",
"Category:Dirac Notation"
] |
proofwiki-23633 | Linear Transform on Dirac Bra | Let $\bra x$ be a bra in some dual space.
Let $T$ be a linear transform.
Then:
:$\bra x T = \bra {T^* x}$ | {{begin-eqn}}
{{eqn | l = \bra x T
| r = \ket x^* T
| c = {{Defof|Dirac Notation/Bra|Bra}}
}}
{{eqn | r = \paren {T^* \ket x}^*
| c = Adjoint of Composition of Linear Transformations is Composition of Adjoints
}}
{{eqn | r = \ket {T^* x}^*
| c = {{Defof|Dirac Notation/Ket/Abuse of Notation|Ket n... | Let $\bra x$ be a [[Definition:Dirac Notation/Bra|bra]] in some [[Definition:Dual Vector Space|dual space]].
Let $T$ be a [[Definition:Linear Transformation|linear transform]].
Then:
:$\bra x T = \bra {T^* x}$ | {{begin-eqn}}
{{eqn | l = \bra x T
| r = \ket x^* T
| c = {{Defof|Dirac Notation/Bra|Bra}}
}}
{{eqn | r = \paren {T^* \ket x}^*
| c = [[Adjoint of Composition of Linear Transformations is Composition of Adjoints]]
}}
{{eqn | r = \ket {T^* x}^*
| c = {{Defof|Dirac Notation/Ket/Abuse of Notation|K... | Linear Transform on Dirac Bra | https://proofwiki.org/wiki/Linear_Transform_on_Dirac_Bra | https://proofwiki.org/wiki/Linear_Transform_on_Dirac_Bra | [
"Bras",
"Dirac Notation"
] | [
"Definition:Dirac Notation/Bra",
"Definition:Algebraic Dual/Vector Space",
"Definition:Linear Transformation"
] | [
"Adjoint of Composition of Linear Transformations is Composition of Adjoints",
"Category:Bras",
"Category:Dirac Notation"
] |
proofwiki-23634 | Dirac Inner Product is Conjugate Linear in First Argument | Let $\ket \alpha$, $\ket \beta$, and $\ket \gamma$ be kets.
Let $k$ be a scalar.
Then:
:$\braket {k \alpha + \beta} \gamma = \bar k \braket \alpha \gamma + \braket \beta \gamma$ | We have:
{{begin-eqn}}
{{eqn | l = \braket {k \alpha + \beta} \gamma
| r = \overline {\braket \gamma {k \alpha + \beta} }
| c = Inner product is conjugate symmetric
}}
{{eqn | r = \overline {k \braket \gamma \alpha + \braket \gamma \beta}
| c = Dirac inner product is linear in the second argument
}}
{... | Let $\ket \alpha$, $\ket \beta$, and $\ket \gamma$ be [[Definition:Ket|kets]].
Let $k$ be a [[Definition:Scalar|scalar]].
Then:
:$\braket {k \alpha + \beta} \gamma = \bar k \braket \alpha \gamma + \braket \beta \gamma$ | We have:
{{begin-eqn}}
{{eqn | l = \braket {k \alpha + \beta} \gamma
| r = \overline {\braket \gamma {k \alpha + \beta} }
| c = [[Definition:Inner Product|Inner product]] is [[Definition:Conjugate Symmetric Mapping|conjugate symmetric]]
}}
{{eqn | r = \overline {k \braket \gamma \alpha + \braket \gamma \bet... | Dirac Inner Product is Conjugate Linear in First Argument | https://proofwiki.org/wiki/Dirac_Inner_Product_is_Conjugate_Linear_in_First_Argument | https://proofwiki.org/wiki/Dirac_Inner_Product_is_Conjugate_Linear_in_First_Argument | [
"Dirac Notation",
"Inner Products",
"Quantum Mechanics"
] | [
"Definition:Dirac Notation/Ket",
"Definition:Scalar"
] | [
"Definition:Inner Product",
"Definition:Conjugate Symmetric Mapping",
"Definition:Dirac Notation/Inner Product",
"Definition:Linear Transformation",
"Sum of Complex Conjugates",
"Product of Complex Conjugates",
"Definition:Inner Product",
"Definition:Conjugate Symmetric Mapping",
"Category:Dirac Not... |
proofwiki-23635 | Power Series Expansion for Logarithm of m over n | {{begin-eqn}}
{{eqn | l = \map \ln {\frac m n}
| r = 2 \sum_{n \mathop = 0}^\infty \frac 1 {2 n + 1} \paren {\frac {m - n} {m + n} }^{2 n + 1}
}}
{{eqn | r = 2 \paren {\frac {m - n} {m + n} + \frac 1 3 \paren {\frac {m - n} {m + n} }^3 + \frac 1 5 \paren {\frac {m - n} {m + n} }^5 + \frac 1 7 \paren {\frac {m - n... | From Power Series Expansion for $\map \ln {\dfrac {1 + x} {1 - x} }$:
:$(1): \quad \ds \map \ln {\dfrac {1 + x} {1 - x} } = 2 \paren {x + \frac {x^3} 3 + \frac {x^5} 5 + \frac {x^7} 7 + \cdots}$
for $-1 < x \le 1$.
Let $m, n \in \R_{>0}$.
Let $x = \dfrac {m - n} {m + n}$.
Then as $m$ and $n$ are both positive:
:$0 < x ... | {{begin-eqn}}
{{eqn | l = \map \ln {\frac m n}
| r = 2 \sum_{n \mathop = 0}^\infty \frac 1 {2 n + 1} \paren {\frac {m - n} {m + n} }^{2 n + 1}
}}
{{eqn | r = 2 \paren {\frac {m - n} {m + n} + \frac 1 3 \paren {\frac {m - n} {m + n} }^3 + \frac 1 5 \paren {\frac {m - n} {m + n} }^5 + \frac 1 7 \paren {\frac {m - n... | From [[Power Series Expansion for Half Logarithm of 1 + x over 1 - x|Power Series Expansion for $\map \ln {\dfrac {1 + x} {1 - x} }$]]:
:$(1): \quad \ds \map \ln {\dfrac {1 + x} {1 - x} } = 2 \paren {x + \frac {x^3} 3 + \frac {x^5} 5 + \frac {x^7} 7 + \cdots}$
for $-1 < x \le 1$.
Let $m, n \in \R_{>0}$.
Let $x = \df... | Power Series Expansion for Logarithm of m over n | https://proofwiki.org/wiki/Power_Series_Expansion_for_Logarithm_of_m_over_n | https://proofwiki.org/wiki/Power_Series_Expansion_for_Logarithm_of_m_over_n | [
"Power Series Expansion for Logarithm of m over n",
"Examples of Power Series",
"Logarithms"
] | [] | [
"Power Series Expansion for Half Logarithm of 1 + x over 1 - x",
"Definition:Positive/Real Number"
] |
proofwiki-23636 | Sum from 1 to Infinity of x^r over r(r+1) | {{begin-eqn}}
{{eqn | l = \sum_{r \mathop = 1}^\infty \dfrac {x^r} {r \paren {r + 1} }
| r = \dfrac x {1 \times 2} + \dfrac {x^2} {2 \times 3} + \dfrac {x^3} {3 \times 4} + \cdots
| c =
}}
{{eqn | r = \paren {\dfrac 1 x - 1} \map \ln {1 - x} + 1
| c =
}}
{{end-eqn}}
valid for $-1 \le x < 1$. | Let:
:$S = \ds \sum_{r \mathop = 1}^\infty \dfrac {x^r} {r \paren {r + 1} }$
From Partial Fractions Expansion: $\dfrac 1 {x \paren {x + 1} }$:
:$\dfrac 1 {r \paren {r + 1} } = \dfrac 1 r - \dfrac 1 {r + 1}$
and so:
:$\dfrac {x^r} {r \paren {r + 1} } = \dfrac {x^r} r - \dfrac {x^r} {r + 1}$
Hence:
{{begin-eqn}}
{{eqn | ... | {{begin-eqn}}
{{eqn | l = \sum_{r \mathop = 1}^\infty \dfrac {x^r} {r \paren {r + 1} }
| r = \dfrac x {1 \times 2} + \dfrac {x^2} {2 \times 3} + \dfrac {x^3} {3 \times 4} + \cdots
| c =
}}
{{eqn | r = \paren {\dfrac 1 x - 1} \map \ln {1 - x} + 1
| c =
}}
{{end-eqn}}
valid for $-1 \le x < 1$. | Let:
:$S = \ds \sum_{r \mathop = 1}^\infty \dfrac {x^r} {r \paren {r + 1} }$
From [[Partial Fractions Expansion/Examples/1 over x(x+1)|Partial Fractions Expansion: $\dfrac 1 {x \paren {x + 1} }$]]:
:$\dfrac 1 {r \paren {r + 1} } = \dfrac 1 r - \dfrac 1 {r + 1}$
and so:
:$\dfrac {x^r} {r \paren {r + 1} } = \dfrac {x^... | Sum from 1 to Infinity of x^r over r(r+1) | https://proofwiki.org/wiki/Sum_from_1_to_Infinity_of_x^r_over_r(r+1) | https://proofwiki.org/wiki/Sum_from_1_to_Infinity_of_x^r_over_r(r+1) | [
"Examples of Power Series"
] | [] | [
"Partial Fractions Expansion/Examples/1 over x(x+1)",
"Translation of Index Variable of Summation",
"Power Series Expansion for Logarithm of 1 - x"
] |
proofwiki-23637 | Polarization Identity/Complex Vector Space/Corollary | Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space over $\C$.
Let $\norm \cdot$ be the inner product norm on $V$.
Then, we have:
:$4 \innerprod x y = \norm {x + y}^2 - \norm {x - y}^2 + i \norm {x + i y}^2 - i \norm {x - iy}^2$
for each $x, y \in V$. | From Inner Product is Sesquilinear, the mapping $q : V \times V \to \C$ defined by:
:$\map q {x, y} = \innerprod x y$ for each $\tuple {x, y} \in V \times V$
is sesquilinear.
Define $Q : V \to \R$ by:
:$\map Q x = \innerprod x x$ for each $x \in V$.
Let $\norm {\, \cdot \,}$ be the inner product norm of $\innerprod \cd... | Let $\struct {V, \innerprod \cdot \cdot}$ be an [[Definition:Inner Product Space|inner product space]] over $\C$.
Let $\norm \cdot$ be the [[Definition:Inner Product Norm|inner product norm]] on $V$.
Then, we have:
:$4 \innerprod x y = \norm {x + y}^2 - \norm {x - y}^2 + i \norm {x + i y}^2 - i \norm {x - iy}^2$... | From [[Inner Product is Sesquilinear]], the [[Definition:Mapping|mapping]] $q : V \times V \to \C$ defined by:
:$\map q {x, y} = \innerprod x y$ for each $\tuple {x, y} \in V \times V$
is [[Definition:Sesquilinear Form|sesquilinear]].
Define $Q : V \to \R$ by:
:$\map Q x = \innerprod x x$ for each $x \in V$.
Let $\no... | Polarization Identity/Complex Vector Space/Corollary | https://proofwiki.org/wiki/Polarization_Identity/Complex_Vector_Space/Corollary | https://proofwiki.org/wiki/Polarization_Identity/Complex_Vector_Space/Corollary | [
"Polarization Identity"
] | [
"Definition:Inner Product Space",
"Definition:Inner Product Norm"
] | [
"Inner Product is Sesquilinear",
"Definition:Mapping",
"Definition:Sesquilinear Form",
"Definition:Inner Product Norm",
"Polarization Identity/Complex Vector Space",
"Category:Polarization Identity"
] |
proofwiki-23638 | Densely-Defined Linear Operator with Real-Valued Quadratic Form is Symmetric | Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space.
Let $\struct {\map D T, T}$ be a densely-defined linear operator such that:
:$\innerprod {T x} x \in \R$ for each $x \in \map D T$.
Then $T$ is symmetric. | Define $q : \map D T \times \map D T \to \C$ by:
:$\map q {x, y} = \innerprod {T x} y$ for each $\tuple {x, y} \in \map D T \times \map D T$.
To show that $T$ is symmetric, we establish that:
:$\map q {x, y} = \overline {\map q {y, x} }$ for each $\tuple {x, y} \in \map D T \times \map D T$.
This will show that:
:$\inn... | Let $\struct {\HH, \innerprod \cdot \cdot}$ be a [[Definition:Hilbert Space|Hilbert space]].
Let $\struct {\map D T, T}$ be a [[Definition:Densely-Defined Linear Operator|densely-defined linear operator]] such that:
:$\innerprod {T x} x \in \R$ for each $x \in \map D T$.
Then $T$ is [[Definition:Symmetric Densely-De... | Define $q : \map D T \times \map D T \to \C$ by:
:$\map q {x, y} = \innerprod {T x} y$ for each $\tuple {x, y} \in \map D T \times \map D T$.
To show that $T$ is [[Definition:Symmetric Densely-Defined Linear Operator|symmetric]], we establish that:
:$\map q {x, y} = \overline {\map q {y, x} }$ for each $\tuple {x, y} ... | Densely-Defined Linear Operator with Real-Valued Quadratic Form is Symmetric | https://proofwiki.org/wiki/Densely-Defined_Linear_Operator_with_Real-Valued_Quadratic_Form_is_Symmetric | https://proofwiki.org/wiki/Densely-Defined_Linear_Operator_with_Real-Valued_Quadratic_Form_is_Symmetric | [
"Densely-Defined Linear Operators",
"Symmetric Densely-Defined Linear Operators"
] | [
"Definition:Hilbert Space",
"Definition:Densely-Defined Linear Operator",
"Definition:Symmetric Densely-Defined Linear Operator"
] | [
"Definition:Symmetric Densely-Defined Linear Operator",
"Polarization Identity/Complex Vector Space",
"Definition:Linear Transformation",
"Inner Product is Sesquilinear",
"Sum of Complex Conjugates",
"Product of Complex Conjugates",
"Category:Densely-Defined Linear Operators",
"Category:Symmetric Dens... |
proofwiki-23639 | Positive Linear Operator is Symmetric | Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space.
Let $\struct {\map D T, T}$ be a positive linear operator.
Then $T$ is symmetric. | From the definition of a positive linear operator, we have:
:$\innerprod {T x} x \in \R$ for each $x \in \map D T$.
Hence, from Densely-Defined Linear Operator with Real-Valued Quadratic Form is Symmetric, we have that $T$ is symmetric.
{{qed}}
Category:Symmetric Densely-Defined Linear Operators
b5zozm4pd19qllk82jw5e8p... | Let $\struct {\HH, \innerprod \cdot \cdot}$ be a [[Definition:Hilbert Space|Hilbert space]].
Let $\struct {\map D T, T}$ be a [[Definition:Positive Linear Operator|positive linear operator]].
Then $T$ is [[Definition:Symmetric Densely-Defined Linear Operator|symmetric]]. | From the definition of a [[Definition:Positive Linear Operator|positive linear operator]], we have:
:$\innerprod {T x} x \in \R$ for each $x \in \map D T$.
Hence, from [[Densely-Defined Linear Operator with Real-Valued Quadratic Form is Symmetric]], we have that $T$ is [[Definition:Symmetric Densely-Defined Linear Ope... | Positive Linear Operator is Symmetric | https://proofwiki.org/wiki/Positive_Linear_Operator_is_Symmetric | https://proofwiki.org/wiki/Positive_Linear_Operator_is_Symmetric | [
"Symmetric Densely-Defined Linear Operators"
] | [
"Definition:Hilbert Space",
"Definition:Positive Linear Operator",
"Definition:Symmetric Densely-Defined Linear Operator"
] | [
"Definition:Positive Linear Operator",
"Densely-Defined Linear Operator with Real-Valued Quadratic Form is Symmetric",
"Definition:Symmetric Densely-Defined Linear Operator",
"Category:Symmetric Densely-Defined Linear Operators"
] |
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